The mean of the Beta distribution is

$$\mu = \frac {\alpha}{\alpha + \beta}$$

We want to see whether restricting the permissible range of $\mu$, will guarantee that we will have either $\{\alpha \geq 1, \beta >0\}$, OR $\{\alpha >0, \beta \geq 1\}$.

Treating the mean as a function of the parameters we obtain

$$\frac {\partial \mu}{\partial \alpha} > 0, \;\; \frac {\partial \mu}{\partial \beta} <0$$

So it is monotonically increasing in $\alpha$ and monotonically decreasing in $\beta$.

So

$$\min \mu = 1/(1+\beta)\implies \{\alpha \geq 1, \beta >0\} \tag {1}$$

but the situation $\{\alpha >0, \beta \geq 1\}$ permits all possible values of $\mu$ (in $(0,1)$).

In other words by restricting the mean to lie in the interval $[1/(1+\beta),\, 1)$ we can guarantee that we will have $\{\alpha \geq 1, \beta >0\}$. But there is no restriction on the mean that will guarantee us that $\{\alpha >0, \beta \geq 1\}$.

So we should turn to the variance, which is

$$\sigma^2 = \frac {\alpha \beta}{(\alpha + \beta)^2(\alpha + \beta +1)}$$

It is not difficult to determine that no restriction on the range of the variance can guarantee that we will have $\{\alpha >0, \beta \geq 1\}$.

So the answer to the question is that the best we can do is to guarantee what relation $(1)$ reflects, by restricting the value of the mean from below.

The mean of the Beta distribution is

$$\mu = \frac {\alpha}{\alpha + \beta}$$

We want to see whether restricting the permissible range of $\mu$, will guarantee that we will have either $\{\alpha \geq 1, \beta >0\}$, OR $\{\alpha >0, \beta \geq 1\}$.

Treating the mean as a function of the parameters we obtain

$$\frac {\partial \mu}{\partial \alpha} > 0, \;\; \frac {\partial \mu}{\partial \beta} <0$$

So it is monotonically increasing in $\alpha$ and monotonically decreasing in $\beta$.

So

$$\min \mu = 1/(1+\beta)\implies \{\alpha \geq 1, \beta >0\} \tag {1}$$

but the situation $\{\alpha >0, \beta \geq 1\}$ permits all possible values of $\mu$ (in $(0,1)$).

In other words by restricting the mean to lie in the interval $[1/(1+\beta),\, 1)$ we can guarantee that we will have $\{\alpha \geq 1, \beta >0\}$. But there is no restriction on the mean that will guarantee us that $\{\alpha >0, \beta \geq 1\}$.

So we should turn to the variance, which is

$$\sigma^2 = \frac {\alpha \beta}{(\alpha + \beta)^2(\alpha + \beta +1)}$$

It is not difficult to determine that no restriction on the range of the variance can guarantee that we will have $\{\alpha >0, \beta \geq 1\}$.

So :

If we impose the restriction $\mu \geq 1/(1+\beta)$, then we will certainly have $\{\alpha \geq 1, \beta >0\}$, i.e. "not both parameters smaller than unity".

But this in a sense is a partial result, since there is also the other way in which "not both parameters are smaller than unity". In other words, this approach imposes the additional restriction that only $\beta$ is allowed to be smaller than unity. It is therefore incompatible with Beta distributions for which we want to be able to have $\alpha \leq 1$ (and $\beta >1$).

The mean of the Beta distribution is

$$\mu = \frac {\alpha}{\alpha + \beta}$$

We wan tot see whether restricting the permissible range of $\mu$, will guarantee that we will have either $\{\alpha \geq 1, \beta >0\}$, OR $\{\alpha >0, \beta \geq 1\}$.

Treating the mean as a function of the parameters we obtain

$$\frac {\partial \mu}{\partial \alpha} > 0, \;\; \frac {\partial \mu}{\partial \beta} <0$$

So it is monotonically increasing in $\alpha$ and monotonically decreasing in $\beta$.

So

$$\min \mu = 1/(1+\beta)\implies \{\alpha \geq 1, \beta >0\} \tag {1}$$

but the situation $\{\alpha >0, \beta \geq 1\}$ permits all possible values of $\mu$ (in $(0,1)$).

In other words by restricting the mean to lie in the interval $[1/(1+\beta),\, 1)$ we can guarantee that we will have $\{\alpha \geq 1, \beta >0\}$. But there is no restriction on the mean that will guarantee us that $\{\alpha >0, \beta \geq 1\}$.

So we should turn to the variance, which is

$$\sigma^2 = \frac {\alpha \beta}{(\alpha + \beta)^2(\alpha + \beta +1)}$$

It is not difficult to determine that no restriction on the range of the variance can guarantee that we will have $\{\alpha >0, \beta \geq 1\}$.

So the answer to the question is that the best we can do is to guarantee what relation $(1)$ reflects, by restricting the value of the mean from below.

The mean of the Beta distribution is

$$\mu = \frac {\alpha}{\alpha + \beta}$$

We want to see whether restricting the permissible range of $\mu$, will guarantee that we will have either $\{\alpha \geq 1, \beta >0\}$, OR $\{\alpha >0, \beta \geq 1\}$.

Treating the mean as a function of the parameters we obtain

$$\frac {\partial \mu}{\partial \alpha} > 0, \;\; \frac {\partial \mu}{\partial \beta} <0$$

So it is monotonically increasing in $\alpha$ and monotonically decreasing in $\beta$.

So

$$\min \mu = 1/(1+\beta)\implies \{\alpha \geq 1, \beta >0\} \tag {1}$$

but the situation $\{\alpha >0, \beta \geq 1\}$ permits all possible values of $\mu$ (in $(0,1)$).

In other words by restricting the mean to lie in the interval $[1/(1+\beta),\, 1)$ we can guarantee that we will have $\{\alpha \geq 1, \beta >0\}$. But there is no restriction on the mean that will guarantee us that $\{\alpha >0, \beta \geq 1\}$.

So we should turn to the variance, which is

$$\sigma^2 = \frac {\alpha \beta}{(\alpha + \beta)^2(\alpha + \beta +1)}$$

It is not difficult to determine that no restriction on the range of the variance can guarantee that we will have $\{\alpha >0, \beta \geq 1\}$.

So the answer to the question is that the best we can do is to guarantee what relation $(1)$ reflects, by restricting the value of the mean from below.